Question

A golfer tees off from the top of a rise, giving the golf ball an initial velocity of 44 m/s at an angle of 25° above the horizontal. The ball strikes the fairway a horizontal distance of 190 m from the tee. Assume the fairway is level. (a) How high is the rise above the fairway? (b) What is the speed of the ball as it strikes the fairway?

Answer 1

Answer:

a) $y=17.6m$

b) $v_{x}=39.87m/s\\v_{y}=-28.05m/s$

Explanation:

From the exercise we got our initial data

$v_{o}=44m/s\\\beta =25\\x=190m$

a) To find maximum height we know that at that point $v_{y}=0$

$v_{y} ^{2} =v_{oy} ^{2} +2a(y-y_{o} )$

$0=(44sin25)^{2} -2(9.8)y$

Solving for y

$y=\frac{(44sin25)^{2} }{2(9.8)} =17.6 m/s$

b) Since we know that the ball strikes the fairway 190 m away

$x=v_{ox}t\\ 190=44cos25t$

Solving for t

$t=4.76s$

Now, we can calculate the speed of the ball in both axes

$v_{x}=44cos25=39.87m/s$

$v_{y}=v_{oy}+at$

$v_{y}=44sin25-(9.8)(4.76)=-28.05m/s$

The negative sign means the direction of the ball at that point.